CAT 2023 Slot 2 Solved Paper

Let the first term of both series be ${\mathit{a}}_1$, and ${\mathit{b}}_1$, respectively, and the common difference be ${\mathit{d}}_1$, and ${\mathit{d}}_2$, respectively.
It is given that ${\mathit{a}}_5={\mathit{b}}_9$, which implies ${\mathit{a}}_1+4{\mathit{d}}_1={\mathit{b}}_1+8{\mathit{d}}_2$
$\Rightarrow {\mathit{a}}_1-{\mathit{b}}_1=8{\mathit{d}}_2-4{\mathit{d}}_1.....\mathrm{Eq}\left(1\right)$
Similarly, it is known that a_ $\left\{19\right\}={\mathrm{b}}_{-}\left\{19\right\}$, which implies ${\mathit{a}}_1+18{\mathit{d}}_1={\mathit{b}}_1+18{\mathit{d}}_2$
$\Rightarrow {\mathit{a}}_1-{\mathit{b}}_1=18{\mathit{d}}_2-18{\mathit{d}}_1.....\mathrm{Eq}\left(2\right)$
Equating (1) and (2), we get:
$\Rightarrow 18{\mathit{d}}_2-18{\mathit{d}}_1=8{\mathit{d}}_2-4{\mathit{d}}_1$
$\Rightarrow 10{\mathit{d}}_2=14{\mathit{d}}_1$
$\Rightarrow 5{\mathit{d}}_2=7{\mathit{d}}_1$
Since, ${\mathit{d}}_1,{\mathit{d}}_2$ are the prime numbers, which implies ${\mathit{d}}_1=5,{\mathit{d}}_2=7$.
It is also known that ${\mathit{b}}_2=0$, which implies ${\mathit{b}}_1+{\mathit{d}}_2=0\Rightarrow {\mathit{b}}_1=-{\mathit{d}}_2=-7$
Putting the value of ${\mathit{b}}_1,{\mathit{d}}_1$, and, ${\mathit{d}}_2$ in $\mathrm{Eq}\left(1\right)$, we get:

Hence, The correct option is B